Answer :
Final Answer:
The age of the rock containing 18.3 mg of 238U and 1.69 mg of 206Pb is approximately 1.23 billion years.
Explanation:
To determine the age of the rock, we can use the radioactive decay of uranium-238 (238U) to lead-206 (206Pb). The half-life of 238U is given as 4.51 × 10^9 years.
Calculate the ratio of the amount of 206Pb to 238U:
1.69 mg / 18.3 mg = 0.0921
Take the natural logarithm of the ratio:
ln(0.0921) ≈ -2.385
Calculate the age using the formula for radioactive decay:
Age = (ln(ratio) / -0.693) * (t1/2 of 238U)
Age ≈ (-2.385 / -0.693) * (4.51 × 10^9 years)
Age ≈ 1.23 billion years
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Final answer:
The age of the rock that contains 18.3 mg of 238U and 1.69 mg of 206Pb can be determined using the radioactive decay equation. By calculating the ratio of 206Pb to 238U and using the half-life of 238U, we can find the age of the rock. The calculated age will depend on the specific values of the ratio and the half-life.
Explanation:
To determine the age of the rock, we can use the radioactive decay equation and the given amounts of 238U and 206Pb. The decay equation for 238U is:
238U → 206Pb + α particles
First, we need to calculate the ratio of 238U to 206Pb in the rock sample. Let's call this ratio R:
R = (amount of 206Pb) / (amount of 238U)
Using the given values:
R = 1.69 mg / 18.3 mg = 0.0923
Next, we can use the half-life of 238U to calculate the age of the rock. The decay equation can be expressed as:
t = (ln(R) / ln(0.5)) * (half-life of 238U)
Substituting the values:
t = (ln(0.0923) / ln(0.5)) * (4.51 × 10^9 years)
Calculating this expression gives us the age of the rock.
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